Complex Numbers And Quadratic Equations question 561
Question: If $ \alpha ,\beta $ be the roots of the equation $ 2x^{2}-35x+2=0 $ then the value of $ {{(2\alpha -35)}^{3}}.{{(2\beta -35)}^{3}} $ is equal to [Bihar CEE 1994]
Options:
A) 1
B) 64
C) 8
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Since $ \alpha ,\beta $ are the roots of the equation $ 2x^{2}-35x+2=0 $ . Also $ \alpha \beta =1 $
$ \therefore 2{{\alpha }^{2}}-35\alpha =-2 $ or $ 2\alpha -35=\frac{-2}{\alpha } $ $ 2{{\beta }^{2}}-35\beta =-2 $ or $ 2\beta -35=\frac{-2}{\beta } $ Now $ {{(2\alpha -35)}^{3}}{{(2\beta -35)}^{3}}={{( \frac{-2}{\alpha } )}^{3}}{{( \frac{-2}{\beta } )}^{3}} $ $ =\frac{8.8}{{{\alpha }^{3}}{{\beta }^{3}}}=\frac{64}{1}=64 $
BETA
